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Thread: Inequality

  1. #1
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    Inequality

    Hi !

    Either $\displaystyle x,y,z> 0$ , $\displaystyle x + y + z = 1$ . Prove That :

    $\displaystyle (1+ \frac{1}{x} ) (1+\frac{1}{y} ) (1+\frac{1}{z} ) \geq 64$
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  2. #2
    Junior Member
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    Hi !

    So, This is My Solution :

    $\displaystyle (1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z}) \geq 64$
    $\displaystyle \Longleftrightarrow$ $\displaystyle (x+1)(y+1)(z+1) \geq 64xyz$
    $\displaystyle \Longleftrightarrow$ $\displaystyle xy+yz+zx+2 \geq 63xyz$
    by AM-GM :
    $\displaystyle xyz \leq \frac{(x+y+z)^3}{27}$
    $\displaystyle \Longleftrightarrow$ $\displaystyle 54xyz \leq 2$ $\displaystyle (1)$
    And We Have : $\displaystyle (x+y+z)( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} ) \geq 9$
    $\displaystyle \Longleftrightarrow$ $\displaystyle xy+yz+zx \geq 9xyz$ $\displaystyle (2)$
    by add $\displaystyle (1)$ and $\displaystyle (2)$ We Have The result sought !
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  3. #3
    MHF Contributor

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    Very good.
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